Theorem

Let H1, H2, ... Hn be subgroups of Group G with operation ${\displaystyle \ast }$

${\displaystyle H_{1}\cap H_{2}\cap \cdots \cap H_{n}}$ with ${\displaystyle \ast }$ is a subgroup of Group G

${\displaystyle \color {RawSienna}(H_{1}\cap H_{2})\subseteq G}$

 1. ${\displaystyle H_{1}\subseteq G}$ H1 is subgroup of G 2. ${\displaystyle H_{2}\subseteq G}$ H2 is subgroup of G 3. ${\displaystyle (H_{1}\cap H_{2})\subseteq G}$ 1. and 2.

${\displaystyle \color {RawSienna}H_{1}\cap H_{2}}$ with ${\displaystyle \color {RawSienna}\ast }$ is a Group

Closure

 4. Choose ${\displaystyle x,y\in (H_{1}\cap H_{2})}$ 5. ${\displaystyle x\ast y\in H_{1}}$ closure of H1 6. ${\displaystyle x\ast y\in H_{2}}$ closure of H2 7. ${\displaystyle x\ast y\in (H_{1}\cap H_{2})}$ 5. and 6.

Associativity

 8. ${\displaystyle \ast }$ is associative on G. Group G's operation is ${\displaystyle \ast }$ 9. ${\displaystyle (H_{1}\cap H_{2})\subseteq G}$ 3. 10. ${\displaystyle \ast }$ is associative on ${\displaystyle (H_{1}\cap H_{2})}$ 8. and 9.

Identity

 11. ${\displaystyle e_{G}\in H_{1}}$ and ${\displaystyle e_{G}\in H_{2}}$ Subgroup H1 and H2 inherit identity from G 12. ${\displaystyle \forall g\in G:e_{G}\ast g=g\ast e_{G}=g}$ eG is identity of G, 13. ${\displaystyle \forall \;g\in (H_{1}\cap H_{2}):e_{G}\ast g=g\ast e_{G}=g}$ ${\displaystyle (H_{1}\cap H_{2})\subseteq G}$ and 9. 14. ${\displaystyle (H_{1}\cap H_{2})}$ has identity eG definition of identity

Inverse

 15. Choose ${\displaystyle g\in (H_{1}\cap H_{2})\subseteq G}$ 16. ${\displaystyle g\in H_{1}}$, ${\displaystyle g\in H_{2}}$, and ${\displaystyle g\in G}$ 17. gH1−1 in H1, and gH2−1 in H2. G, H1, and H2 are groups 18. ${\displaystyle g_{H1}^{-1}\in G}$ ${\displaystyle H_{1}\subseteq G}$ 19. ${\displaystyle g_{H1}^{-1}\ast g=g\ast g_{H1}^{-1}=e_{G}}$ G and H1 shares identity e 20. gH1−1 is inverse of g in G 19. and definition of inverse 21. Let gG−1 be inverse of g has in G 22. gG−1 = gH1−1 inverse is unique 22. gG−1 = gH2−1 similar to 21. 23. ${\displaystyle g^{-1}=g_{H1}^{-1}=g_{H2}^{-1}\in (H_{1}\cap H_{2})}$ 24. g has inverse g−1 in ${\displaystyle (H_{1}\cap H_{2})}$